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JEE Main & Advanced Mathematics Functions Question Bank Functions

  • question_answer
    Range of the function \[f(x)=\frac{{{x}^{2}}}{{{x}^{2}}+1}\]is [Orissa JEE 2005]

    A)            (?1, 0)

    B)            (?1, 1)

    C)            [0, 1)

    D)            (1, 1)

    Correct Answer: C

    Solution :

               Let \[y=\frac{{{x}^{2}}}{{{x}^{2}}+1}\]                    Þ \[(y-1){{x}^{2}}+0x+y=1,y\ne 1\] for real values of x,                    we have \[D\ge 0\Rightarrow -4y(y-1)\ge 0\Rightarrow y(y-1)\le 0\Rightarrow y\in [0,\,1)\]                    \[0\le \frac{{{x}^{2}}}{{{x}^{2}}+1}<1\].


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